Stable solitons of the cubic-quintic NLS with a delta-function potential

Transcription

1 Stable solitons of the cubic-quintic NLS with a delta-function potential François Genoud TU Delft Besançon, 7 January 015

2 The cubic-quintic NLS with a δ-potential We consider the nonlinear Schrödinger equation iψ z + ψ xx + ɛδ(x)ψ + ψ ψ ψ 4 ψ = 0, (NLS) for ψ = ψ(x, z) : R R C. This combination of nonlinearities is well known in optical media, including liquid waveguides and speciality glasses. The delta-function models the interaction of a broad beam with a narrow trapping potential. The parameter ɛ > 0 is a coupling constant which gives the strength of the potential. The Cauchy problem for (NLS) is globally well-posed in H 1 (R).

3 Literature Fukuizumi Ohta Ozawa Jeanjean Fukuizumi Le Coz Fukuizumi Fibich Ksherim Sivan Adami Noja

4 Solitons We look for bound states of the form ψ(x, z) = e ikz u(x), with k > 0, u H 1 (R), u(x) > 0. They must satisfy u ku + ɛδ(x)u + u 3 u 5 = 0, (SNLS) where stands for differentiation with respect to x. More precisely, they satisfy: (i) u ku + u 3 u 5 = 0, x 0; (ii) u C (R \ {0}) C(R); (iii) u is even on R; (iv) u (0 ± ) = ɛ u(0); (v) u(x), u (x) 0 as x.

5 Brute force integration The first integral of (SNLS) reads (u ) (x) ku (x) + u 4 (x) 1 3 u6 (x) = 0, x 0. Taking the limit x 0 ± yields u 6 (0) 3u 4 (0) + 3ku (0) 3(u ) (0) = 0

6 Brute force integration The first integral of (SNLS) reads (u ) (x) ku (x) + u 4 (x) 1 3 u6 (x) = 0, x 0. Taking the limit x 0 ± yields u 6 (0) 3u 4 (0) + 3ku (0) 3 ɛ 4 u (0) = 0

7 Brute force integration The first integral of (SNLS) reads (u ) (x) ku (x) + u 4 (x) 1 3 u6 (x) = 0, x 0. Taking the limit x 0 ± and dividing through by u (0) yields the solutions of which are u 4 (0) 3u (0) + 3(k ɛ 4 ) = 0, u ±,k,ɛ (0) = 3 ( ( ) ) 1 ± k ɛ 4. These numbers exist are positive if and only if ɛ 4 < k ɛ 4.

8 To integrate further, we express u as u 1 (x) = sgn(x)u(x) 3 u4 (x) u (x) + k, x 0. The positivity conditions (PC) under the square root yield two different regimes: (A) ɛ 4 < k < 3 4 : u +,k,ɛ (0) doesn t satisfy (PC), there is only one soliton, u,k,ɛ; (B) 3 4 < k < ɛ 4 : both u ±,k,ɛ (0) satisfy (PC), there are two solitons, u ±,k,ɛ. We note that regime (A) is void if ɛ 3, so we shall assume 0 < ɛ < 3 from now on. Regime (B) is void when ɛ = 0.

9 Plots 3 u k Figure: For ɛ = 0, plot of u L against k. The solutions u,k,0 are well known and take the simple form k u,k,0 (x) = k3 cosh ( kx ), 0 < k < 3 4.

10 4 1 u k Figure: For ɛ = 0.5 3, plot of u L against k. Two solutions coexist for each given wavenumber k > 3 4, with a fold bifurcation occuring at k ɛ = ɛ 4. We will show that all solutions on the curve are orbitally stable.

11 u = u = u(x) 1 u(x) ) x u = ) x u = u(x) 1 u(x) ) x u = ) x u = u(x) 1 u(x) ) x ) x Figure: Soliton profiles for ɛ =

12 4 1 3 u k Figure: For ɛ = 0.9 3, plot of u L against k.

13 Explicit form of the solutions The solutions in regime (A) (i.e. for ɛ 4 < k < 3 4 ) are given by u,k,ɛ (x) = k 1 + ɛ+ɛ 1+(4k/ɛ 1)(1 4k/3) 4( e k x + k ɛ/) (1 4k/3)(, k ɛ/) ɛ+ɛ 1+(4k/ɛ 1)(1 4k/3) e k x which, at k = 3/4, reduces to 3 1 u,3/4,ɛ (x) = 1 + ɛ 3 ɛ e. 3 x

14 In regime (B) ( 3 4 < k < ɛ 4 ) the expressions are more involved: u ±,k,ɛ (x) = k ( e ) (( k( x c) + e k( x c) k )e k( x c) ( ) ), k 3 1 e k( x c) where the integration constants c = c ±,k,ɛ R can be determined from the respective conditions at x = 0, yielding e kc,k,ɛ = ɛ 4k + ɛ k 4k ɛ 4k + 3 k k 3 + ɛ 4k and e kc+,k,ɛ = ɛ 4k ɛ k + 4k ɛ 4k 3 k k 3 + ɛ 4k.

15 At the fold bifurcation point (k ɛ = ɛ 4 ) the solution takes the more tractable form: u ɛ (x) = ɛ 3 + ɛ cosh( 3 + ɛ x ) + ɛ 3 + ɛ sinh( 3 + ɛ x ). This expression will be useful to study the fold bifurcation.

16 Bifurcation and spectral analysis We shall call the lower curve, respectively the upper curve, the sets S,ɛ = { (k, u,k,ɛ ) : k ( ɛ 4, k)}, S +,ɛ = { (k, u +,k,ɛ ) : k ( 3 4, k)}. We also let F ɛ : R H 1 (R) H 1 (R), F ɛ (k, u) = u + ku ɛδ(x)u u 3 + u 5, so that (SNLS) reads F ɛ (k, u) = 0.

17 Theorem 1 (i) The set S ɛ := S,ɛ {(k ɛ, u ɛ )} S +,ɛ is a smooth curve in R H 1 (R), and bifurcates from zero as k ɛ 4, and from infinity as k 3 4 along S +,ɛ. (ii) The linearized operator D u F ɛ (k, u) : H 1 (R) H 1 (R), D u F ɛ (k, u) = d dx + k ɛδ(x) 6u (x) + 5u 4 (x), is non-singular along S ±,ɛ, and singular at (k, u) = (k ɛ, u ɛ ). (iii) Furthermore, D u F ɛ (k, u) has exactly one negative eigenvalue along S,ɛ ; has no negative eigenvalues along S +,ɛ.

18 Lemma The Schrödinger operator D u F ɛ (k, u) = d + k ɛδ(x) 6u (x) + 5u 4 (x) dx has finitely many (simple) eigenvalues below a continuous spectrum [k, ). Furthermore, (i) D u F ɛ (k, u,k,ɛ ) is an isomorphism for all k ( ɛ 4, k); (ii) D u F ɛ (k, u +,k,ɛ ) is an isomorphism for all k ( 3 4, k); (iii) D u F ɛ (k ɛ, u ɛ ) is singular with ker D u F ɛ (k ɛ, u ɛ ) = span{ u ɛ }. Since u ɛ > 0, zero is the principal eigenvalue of D u F ɛ (k ɛ, u ɛ ). Proof. The first statements are standard properties of Schrödinger operators. Then the analysis boils down to checking if ker D u F ɛ (k, u) is trivial or not, which follows by straightforward ODE arguments.

19 Proposition S ɛ,± are smooth curves of non-degenerate solutions of (SNLS). S,ɛ bifurcates from the point ( ɛ 4, 0) in R H1 (R), and meets S +,ɛ at the point (k ɛ, u ɛ ), where D u F ɛ (k, u) becomes singular. S +,ɛ becomes unbounded as k 3 4 : lim k 3 4 u +,k,ɛ L =. Proof. The smoothness of the curves S,ɛ and S +,ɛ follows from the implicit function theorem in R H 1 (R) and the previous lemma. Since ker D u F ɛ ( ɛ 4, 0) = span { e ɛ x }, zero is a simple eigenvalue of D u F ɛ ( ɛ 4, 0) and local bifurcation from ( ɛ 4, 0) in R H1 (R) follows by the Crandall-Rabinowitz theorem (1971). The last statement follows from Rabinowitz s global bifurcation theorem (1971).

20 Local analysis at the fold We need to check: (1) S,ɛ and S +,ɛ meet smoothly at (k ɛ, u ɛ ); () D u F ɛ (k, u) has exactly one negative eigenvalue along S,ɛ ; (3) this eigenvalue crosses zero with non-zero speed as we pass through (k ɛ, u ɛ ) (so doesn t bounce back). () follows by analytic perturbation theory with respect to ɛ (0, 3), from the case ɛ = 0. Indeed, at ɛ = 0, we have ker ( d dx + k 6u,k,0 + 5u4,k,0) = span{u,k,0 }, and so zero is the second eigenvalue by Sturm s oscillation theory.

21 (1) & (3) can be proved using a reparametrization devised by Crandall and Rabinowitz (1973) of the curve about (k ɛ, u ɛ ), where the parametrization by k breaks down. Upon checking that and codim rge D u F ɛ (k, u ɛ ) = dim ker D u F ɛ (k, u ɛ ) = 1 D k F ɛ (k, u ɛ ) rge D u F ɛ (k, u ɛ ), Crandall Rabinowitz = the solutions of (SNLS) in a neighbourhood of (k, u ɛ ) form a smooth curve, {(k s, u s ) : s ( ε, ε)} R H 1 (R) (for some ε > 0) such that, at s = 0, (k 0, u 0 ) = (k ɛ, u ɛ ) and ( k 0, u 0 ) = (0, u ɛ ).

22 We already know that so dim ker D u F ɛ (k ɛ, u ɛ ) = 1. ker D u F ɛ (k ɛ, u ɛ ) = span{ u ɛ }, Furthermore, the range of the self-adjoint operator D u F ɛ (k, u ɛ ) is characterized by { } rge D u F ɛ (k, u ɛ ) = v L (R) : v u ɛ dx = 0, R so codim rge D u F ɛ (k, u ɛ ) = 1. Finally, F ɛ (k, u) = u + ku ɛδ(x)u u 3 + u 5 = D k F ɛ (k, u ɛ ) = u ɛ rge D u F ɛ (k, u ɛ ) since R u ɛ u ɛ dx = 0 u ɛ u ɛ dx = u ɛ(0) < 0.

23 To show that the first eigenvalue of D u F ɛ (k, u) crosses zero at (k ɛ, u ɛ ) we use the local parametrization. Denoting by µ s the first eigenvalue along the curve, this amounts to showing that µ 0 0. At s = 0 (i.e. at (k ɛ, u ɛ )) the first eigenvalue and eigenfunction µ s, v s satisfy µ 0 = 0, v 0 = u ɛ and v s + k s v s ɛδ(x)v s [6u s 5u 4 s ]v s = µ s v s, s ( ε, ε). Recalling that k 0 = k ɛ, u 0 = u ɛ, differentiating with respect to s at s = 0 yields v 0 + k v 0 4[3 5u ɛ]u ɛ η ɛ [6 5u ɛ]u ɛ v 0 = µ 0 u ɛ.

24 Multiplying both sides by u ɛ, integrating by parts and using D u F ɛ (k, u ɛ ) u ɛ = 0 yields µ 0 = 4 R [5u ɛ 3]u ɛ u ɛ 3 R u ɛ, which can be seen to be positive numerically: 1.5 f(ǫ) ǫ Figure: The graph of f (ɛ) := 0 [5u ɛ 3]u ɛ u ɛ 3

25 Stability Due to the U(1)-invariance of (NLS), the appropriate notion of stability in this context is that of orbital stability. Definition We say that the bound state ψ k (x, z) = e ikz u k (x) is orbitally stable if for all ε > 0 there exists δ > 0 such that for any solution ϕ(x, z) of (NLS) with initial data ϕ(, 0) H 1 (R) there holds ϕ(, 0) u k H 1 δ = inf θ R ϕ(, z) eiθ u k H 1 ε for all z 0.

26 Theorem The whole curve S ɛ consists of orbitally stable bound states. Proof. We can use the general theory of orbital stability exposed by Stephan. (I) We know that the spectrum of D u F ɛ (k, u +,k,ɛ ) is strictly positive, for all k ( 3 4, k), so the upper curve is stable. (II) We also know that, for all k ( ɛ 4, k), D uf ɛ (k, u,k,ɛ ) has exactly one simple negative eigenvalue, is non-singular, and the rest of its spectrum is strictly positive. Hence, to complete the proof, we need only show that d dk u,k,ɛ L > 0 k ( ɛ 4, k).

27 Firstly, for k ( 3 4, k), it follows from the expression u,k,ɛ (x) = k ( e ) (( k( x c) + e k( x c) k )e k( x c) that (thanks to Mathematica) ( ) ), k 3 1 e k( x c) d dk u,k,ɛ L = 3ɛ 3+ɛ 4k 3 k 4k 3 > 0. N.B. Similarly, d dk u +,k,ɛ L = 3ɛ 3+ɛ 4k + 3 k 4k 3 < 0 k ( 3 4, k).

28 For k < 3/4, a straightforward (but painful) calculation using u,k,ɛ (x) = shows that where k 1 + ɛ+ɛ 1+(4k/ɛ 1)(1 4k/3) 4( e k x + k ɛ/) u,k,ɛ L = 3 log ϕ ɛ (k) (1 4k/3)( k ɛ/) ɛ+ɛ 1+(4k/ɛ 1)(1 4k/3) e k x 3ɛ + 3ɛ + (4k ɛ )(3 4k) + ( 3 + k )( k ɛ ) ϕ ɛ(k) := 3ɛ + 3ɛ + (4k ɛ )(3 4k) + ( 3 k )( k ɛ ).

29 Finally, d dk ϕ ɛ(k) = 3 3ɛ + (4k ɛ )(3 4k) + k ( 3 + ɛ ɛ k ) 8 k 3ɛ 3ɛ + (4k ɛ )(3 4k)( + 3ɛ + (4k ɛ )(3 4k) + ( 3 k )( k ɛ )) which is strictly positive since k < 3 4 = 3 + ɛ ɛ k > ( 3 ɛ ) + 3ɛ > 0. The proof is complete.